How Far Does the Camera Fall Before the Skydiver Catches It?

[Vaz17]

Hey guys

need help beginning to solve the following:

A skydiver accidentally drops a camera out of the plane. The diver notices the mistake 3.0s later and dives out of the plane with a downward velocity of 10.0 m/s. The camera experiences free fall(9.8 m/s^2), but the sky diver accelerates downwards at 8.0m/s^2.

How far does the camera fall before the sky diver is able to catch it?

I found how far the camera falls in 3.0s= 44.1 m, and Vf of the camera is 29.4 m/s [down] since Vi is 0. I Don't know really where to go from here, any help will be appreciated!

i think the teacher mentioned it takes 39 seconds, but how would i find that

 

[djeitnstine]

What is true about the displacement when the diver meets the camera?

 

[Vaz17]

displacement would be the same as the camera?

 

[djeitnstine]

precisely

 

[Vaz17]

so i need to do d=motion equation of camera=motion equation of diver?

 

[djeitnstine]

You've got it.

 

[djeitnstine]

If you've done it correctly you should get a quadratic

 

[Vaz17]

ok thank you

but how do i deal with the 10m/s and the 8.0 m/s^2 when deriving the divers motion equation? they can't simply be added together can they?

 

[djeitnstine]

I am assuming that you are using y=v_{i}t+ \frac{1}{2}at^{2} where y is the vertical displacement

 

[Vaz17]

yes.

 

[djeitnstine]

Okay so setting 2 equations equal gives v_{i1}t+ \frac{1}{2}a_{1}t^{2}=v_{i2}t+\frac{1}{2}a_{2}t^{2} which gives the quadratic (v_{i1}-v_{i2})t+(a_{1}-a_{2}t^{2})t^{2}=0

So no magic happens =)

Its up to you to decide which root is relevant

edit I left out my 1/2

 

[Vaz17]

thank you very much kind sir :)

 

[djeitnstine]

You are welcome

 

[Vaz17]

ok so what can i derive from that without t

edit where did you leave out the 1/2

 

[djeitnstine]

1) I placed the 1/2's that you see in the equation.

2) no t's should disappear since your velocities and accelerations are different

 

[Fragment]

A simple approach when comparing different velocities is using a table. Of course, I highly recommend working out the mathematics, but a table should also give you the same answer. Example: Car 1: 1s=30m, 2s=50m...Car 2: 1s=45m, 2s=49m...See where they interesect, which is like wording out a quadratic, so-to-speak.

Regards,

Fragment

 

[Vaz17]

hmmm

 

[Fragment]

The solution seems pretty clear to me from this point, djeit pretty much pointed it out. Either graph your quadratics, solve them algebraically, or use the table method.

Fragment

 

[Vaz17]

ok, I found the time, 4.5s, using the quadratic formula. it seems wrong, and once i have this time, how do i get displacement?

which variables do i plug into the displacement formula? my starting values?

 

[djeitnstine]

rework your quadratic...i got 10.7777s

 

[Vaz17]

ok

and do you have to add the 3.0 additional seconds to your final answer though?
aloso, check out this solution on another board, is it the same? this is the one i used and got 4.5, i didnt know how to do the algebra to solve the quadratic you gave me.

http://www.physicshelpforum.com/phy...rk-help/1412-free-fall-question.html#post3682

 

[djeitnstine]

no i did not add the 3 seconds...i was just showing the answer to the quadratic...remember \frac{-b+/- \sqrt{b^{2}-4ac}}{2a}

 

[Vaz17]

lol i think you misunderstood, i didnt know how to solve the equation you gave on pg 1

 

[Fragment]

Well, put in the numbers and let us see what you get.

 

[Vaz17]

ok:)

 

[Vaz17]

LaTeX Code: (v_{i1}-v_{i2})t+(a_{1}-a_{2}t^{2})t^{2}=0

when i plug in the values for this, i get 37.2...:S

 

[Vaz17]

I don't know what to do after that

 

[djeitnstine]

In that case, you should create a post in the math forum asking how to solve a quadratic equation. Also you did not read my equation carefully as the result is not a single number.

 

[Fragment]

You should fix your LaTeX code though, it would help me read what you plugged in. After that we might be able to see where you struggle.

Regards,

Fragment

 

[Vaz17]

no i know how to solve a quadratic, I am just confused with your equation, i think that's the basis of my whole incorrect solution right now lol.

fragment, that code was for djeitnstine's equation on pg 1.

 

[Vaz17]

ok i got it!

now how do i solve for displacement?

 

[Fragment]

Remember that y is your displacement. What did you do by solving your quadratic?

 

[Vaz17]

i solved for time, a little differently then your formula, check the link in one of my above posts.

 

[Fragment]

And so, where are you now in your problem?

 

[Vaz17]

I need displacement given the time it took for the diver to reach the camera.

 

[Fragment]

Didn't you say you had all solved for (obviously except displacement)?
y=delta-d(displacement)

 

[Vaz17]

yes thank you, i just realized i can now sub in my time in each of the motion equations, the camera and diver.

 

[Vaz17]

y=delta-d(displacement)

can you explaint this formula please? i thought y=displacement? lol

 

[Fragment]

Sorry... I haven't bothered with LaTeX yet, basically what I typed was y=displacement, since the symbol for displacement is delta-d. Djeit's derivation had a mistake in it, try:

<br /> (v_{i1}-v_{i2})t+\frac{1}{2}(a_{1}-a_{2})t^{2}=0<br />

 

[djeitnstine]

Hey I'm so sorry about my quadratic equation :S it was supposed to be (v_{i1}-v_{i2})t+\frac{1}{2}(a_{1}-a_{2})t^{2}=0

 

[Vaz17]

oh ok thank you guys very much

the original screwed me up so bad lol

 

[Fragment]

Since it's for your class, you should be able to do this yourself. Make sure you get a lot of practice, test time won't let you use PF

All we did was simply group the like terms together. Make sure you're comfortable with this, because a question like this is bound to be on one of your tests. (From experience)

Regards,

Fragment

 

[Vaz17]

haha you're right. And yes there may be one of these, for the most part we are doing slightly less difficult ones, without the 3 second start difference.

 

[Vaz17]

what answer did you get with the corrected quadratic?

i think you guys are missing the fact that the camera was falling for 3 seconds already and traveled 44.1 m...and neither of those numbers appeared in the quadratic...wow I am nowhere near close to finishing this lol

 

[Fragment]

I don't think I'm allowed to post an answer until you have first, that way I don't give out an easy solution. Did you manage to figure it out yourself?

Regards,

Fragment

 

[Vaz17]

i don't see why it would hurt, I would only gain from it.

but anyways, i got a time of 50 s from the quadratic, and need displacement still :(

im doing something wrong

 

[Fragment]

Can you post your work so we can see where you went wrong? 50s is not correct. It would be very helpful

Fragment

 

[Vaz17]

I don't know how to do those fancy equations so, this is the best i can do

(0-10)t + 1/2 (9.8-8)t^2=0
-10t + 0.9t^2=0

if you sub that into the quadratic you get 50 or another number if you use the other root...

i also want to know where the 3 second delay of the diver comes into play in these equations...because I am not seeing it

 

[Fragment]

<br /> (v_{i1}-v_{i2})t+\frac{1}{2}(a_{1}-a_{2})t^{2}=0<br />

Your numbers are wrong, that's why the quadratic gives you 50...
v_1 here is the first velocity of the camera, and you had found that yourself in your first post (29.4), so I don't know how you got 0 there. As for a1 and a2, they are your accelerations of the camera and of the diver, respectively. (9.8 and 8.0). Try again with those numbers.



Fragment

 

[Vaz17]

wouldnt initial velocity of the camera be 0?

if v1 is 29.4, what's v2?